Question

Two ships are on opposite shores of a narrow sea channel. At the same time, they start moving toward each other at different, but constant speeds. When they pass each other the first time, they are 7.0 km from one shoreline. They continue to the opposite shore, then turn around and start moving toward each other again. When they pass for the second time, they are 3.0 km from the other shoreline. Ignore any effects from winds, waves, or currents, and don’t worry about the ships accelerating, or stopping at the shorelines. How wide is the sea channel?

Answer 1

By setting up equations based on the distances traveled by the ships and their constant speeds, we find the width of the sea channel to be 14 km.

Step-by-step explanation:

To solve this relative motion problem, we will consider the distances covered by the ships and their speeds. Let's call the width of the channel 'W', the speed of the first ship 'S1', and the speed of the second ship 'S2'. When the ships meet for the first time, ship 1 has traveled 'W - 7 km' and ship 2 has traveled '7 km'.

After turning around, ship 1 will travel '7 km' to reach the opposite shore, and then 'W - 3 km' to meet ship 2 for the second time. Similarly, ship 2 will travel 'W - 7 km' and then '3 km' in total. At their second meeting, ship 1 has traveled 'W + 4 km', and ship 2 has traveled '2W - 4 km'.

Since they travel the same time between meetings and speeds are constant, the ratio of their distances is the ratio of their speeds:
S1/S2 = (W + 4 km) / (2W - 4 km)

For the first meeting:
S1/S2 = (W - 7 km) / (7 km)

Equating the two ratios, we have:
(W - 7 km) / (7 km) = (W + 4 km) / (2W - 4 km)
Cross-multiplying gives us:
W^2 - 4W - 28 km^2 + 8 km = 0

Solving for 'W' with the quadratic formula, we get:
W = (4 + sqrt(4^2 + 4 * 28)) / 2 km
which simplifies to:
W = 14 km

Therefore, the width of the sea channel is 14 km.